1 problem found
Two particles \(A\) and \(B\) of masses \(m\) and \(2 m\), respectively, are connected by a light spring of natural length \(a\) and modulus of elasticity \(\lambda\). They are placed on a smooth horizontal table with \(AB\) perpendicular to the edge of the table, and \(A\) is held on the edge of the table. Initially the spring is at its natural length. Particle \(A\) is released. At a time \(t\) later, particle \(A\) has dropped a distance \(y\) and particle \( B\) has moved a distance \(x\) from its initial position (where \(x < a\)). Show that \( y + 2x= \frac12 gt^2\). The value of \(\lambda\) is such that particle \(B\) reaches the edge of the table at a time \(T\) given by \(T= \sqrt{6a/g\,}\,\). By considering the total energy of the system (without solving any differential equations), show that the speed of particle \(B\) at this time is \(\sqrt{2ag/3\,}\,\).
Solution: \begin{align*} \text{N2}(\downarrow): && mg -T &= m\ddot{y} \\ \text{N2}(\rightarrow): && T &= 2m\ddot{x} \\ \Rightarrow && g &= \ddot{y}+2\ddot{x} \\ \Rightarrow && \tfrac12gt^2 &= y + 2x \end{align*} At time \(T = \sqrt{6a/g}\), we have \(y + 2x = 3a\), note also that \(\dot{y}+2\dot{x} = gt\) \begin{array}{ccc} & \text{KE} & \text{GPE} & \text{EPE} \\ \text{Initial} & 0 & 0 & 0 \\ \text{Final} & \frac12m\dot{y}^2 + \frac12(2m)\dot{x}^2 & -mgy & \frac{\lambda (y-x)^2}{2a} \end{array} Also note when we head over the table, \(x = a\) and \(y = a\) \begin{align*} \text{COE}: && 0 &= \frac12m(gT-2\dot{x})^2+m\dot{x}^2-mga+\frac{\lambda(0)^2}{2a} \\ \Rightarrow && 0 &= (gT-2\dot{x})^2+2\dot{x}^2-2ga \\ &&&= (\sqrt{6ag}-2\dot{x})^2+2\dot{x}^2-2ga \\ &&&= 6\dot{x}^2-4\sqrt{6ag}+4ag \\ \Rightarrow &&&= (\sqrt{6}\dot{x} - 2\sqrt{ag})^2 \\ \Rightarrow && \dot{x} &= \sqrt{2ag/3} \end{align*} as required.